<p>Let <em>f(x)</em> be a continuous function in (0, 1) satisfying \(\int_0^1 x\sqrt{x}\, f(x)(1 - \sqrt{x}\,f(x))\,dx = \dfrac{1}{8}\). Number of solutions of the equation \(f(x) = e^x\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand can be rewritten as x√x·f(x)·(1 - √x·f(x)) = x√x[f(x) - x·f(x)²], which when maximized for fixed x occurs at f(x) = 1/(2√x). The constraint integral forces f(x) to satisfy a specific relationship that can be compared with e^x.
<p><strong>Step 1:</strong> Analyze the integrand g(x) = x√x·f(x)·(1 - √x·f(x)). For fixed x ∈ (0,1), treat this as a function of u = √x·f(x).</p><p><strong>Step 2:</strong> Rewrite: g = x·u(1-u). This is maximized when u = 1/2, i.e., when f(x) = 1/(2√x).</p><p><strong>Step 3:</strong> The maximum value is x·(1/4) = x/4. So ∫₀¹ (x/4)dx = 1/8 ✓</p><p><strong>Step 4:</strong> By the method of Lagrange multipliers (or by noting the integral achieves its maximum), we must have f(x) = 1/(2√x) at the solution.</p><p><strong>Step 5:</strong> Now solve f(x) = e^x, i.e., 1/(2√x) = e^x, or e^x·√x = 1/2.</p><p><strong>Step 6:</strong> Define h(x) = e^x·√x on (0,1). We have h'(x) = e^x·√x + e^x/(2√x) = e^x(√x + 1/(2√x)) > 0, so h is strictly increasing.</p><p><strong>Step 7:</strong> As x → 0⁺: h(x) → 0. At x = 1: h(1) = e > 1/2. By IVT, exactly one solution exists in (0,1).</p><p>∴ Answer: B (1 solution)</p>
Correct Answer: B